Question

A rhombus is inscribed in the region common to the two circles \( x^2 + y^2 - 4x - 12 = 0 \) and \( x^2 - y^2 + 4x - 12 = 0 \) with two of its vertices on the line joining the centres of the circles. The area of rhombus is

• A. \( 8\sqrt{3} \)
• B. \( 4\sqrt{2} \)
• C. \( 16\sqrt{3} \)
• D. \( 16\sqrt{2} \)

Hint:

For symmetric circle problems, always locate centres first to exploit geometry.

Answer 1

The Correct Option is A

Concept: Find centres and radii of circles, then use symmetry to determine rhombus diagonals.

Step 1: Find centres. First circle: \[ x^2 + y^2 - 4x - 12 = 0 \Rightarrow (x-2)^2 + y^2 = 16 \] Centre: \( (2,0) \), radius = 4 Second circle: \[ x^2 + y^2 + 4x - 12 = 0 \Rightarrow (x+2)^2 + y^2 = 16 \] Centre: \( (-2,0) \), radius = 4

Step 2: Common region symmetric about y-axis. Distance between centres = 4

Step 3: Maximum rhombus formed: Diagonal along x-axis = 4 Other diagonal: \[ = 2\sqrt{4^2 - 2^2} = 2\sqrt{12} = 4\sqrt{3} \]

Step 4: Area of rhombus: \[ \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3} \] Final Answer: \[ 8\sqrt{3} \]